ship dimensions (TL1)

[whswhs]

Originally Posted by Agemegos

The general formula for the volume of an ellipsoidal cap is doubtless brutal, but in the special case in which the intersecting plane is perpendicular to one of the principle axes of the ellipsoid it seems to me that the volume of the cap must be in proportion to the volume of the ellipsoid as the volume of a spherical cap the same height is to the volume of a sphere, where the radius of the sphere is equal to the semi-axis that the plane cuts. It's a well-behaved linear transformation. The formula for the volume of a spherical cap is tractable.

Is this providing a formula for estimating the volume of a slice through an ellipsoid that extends from one of its poles to a plane of constant latitude? I can see that that would be useful in estimating the draft of a ship of known burden, by estimating the displacement volume needed to float that burden. It looks as if my copy of Hudson's Manual has a formula for that for a sphere, but the print's too small for the current state of my eyes; I'll have to take a look later under brighter light, and possibly with a magnifying glass.

[whswhs]

Originally Posted by Polydamas

The strakes in the Kyrenia ship, which is about your size, varied from 3.1 to 4.3 cm thick. I think it was mostly Aleppo pine, an article on ResearchGate gives the dry density of Aleppo pine from Macedonia as about 0.55 g/cm^3. A sewn boat usually has a few ribs inside for strength, to keep the cargo out of the bilge water and any sharp pointy cargo off the cords, etc. The mast and anchors and steering oar/oars will add some weight, but an average thickness on the order of 10 cm feels high.

I had been going for 7.5 cm, but that's still about double the thickness you cite. It also sounds as if we're looking at a ship with a super-light frame, not a frameless ship. The cited density of Aleppo pine is close enough to half that of water to make a 50% multiplier seem a reasonable figure, and the articles on the Kyrenia ship suggest that it's almost exactly the size I was aiming at, at least in length.

[Anaraxes]

Originally Posted by whswhs

I'm not sure I'm understanding what you're saying. Are you proposing simply to use top deck length and accept the resulting margin of error?

Sorry for the delay in response. But to answer:

No. I'm pointing out that the bow and stern rakes are usually constant. (Here's where a diagram would help, but no pictures on the forum.) Imagine a vertical line dropped from the point of the bow -- the end of the top deck length. For most ships, the bow angles back somewhat, most often in a straight line above the waterline. So there's a triangle formed by that plumb line, the horizontal, and the bow rake line.

We want to know the length of the hull at the waterline. The plumb line has the top deck to keel height of the ship (one of the figures I recall you saying you sometimes could find, even though draft was scarcely reported.) There's some unknown distance along the base of that triangle, the distance between the plumb line and the start of the horizontal part of the keel.

At the waterline, there's another triangle partway between the top-keel triangle and the point at the bow. That triangle has the same proportions as the top-keel triangle, sharing the same bow rake angle. The base of this smaller triangle is the amount you need to subtract from the top deck length to get to the waterline length. You've got the opposite and adjacent sides of the larger triangle, so you could figure that angle and do the trig. Or, since the triangles are similar, it's just a simple proportion. If you're given the freeboard, then you're good, as that's the height of the smaller triangle. If that's not known, you can still write the amount by which to shorten the length in terms of the (still unknown) draft. Then do the algebra to get both draft terms on the same side and solve.

If you don't know the freeboard or top-keel height, then there's not enough information given to find the draft this way. You need some value for that plumb height to use in the proportion.

Some ships do have plumb bows, especially below the waterline. (See, for example, the Iowa-class battleships from WW II; the top part is raked, but the rake stops above the waterline, after which it's plumb. A bow isn't raked always for hydrodynamic reasons.)

A modern bulbous bow is a whole different problem. (I'd be strongly tempted to do the math as though it weren't there, and then tack on volume equal to a half sphere.)

That's all assuming the waterline length reducing is a factor worth correcting for, given that we're starting off by guessing that prismatic coefficient. That limits the error probably to no better than 10-20% anyway.

[DanHoward]

The Ulu Burun shipwreck is the best example of a TL1 trading ship.
http://cgs.la.psu.edu/documents/Pulak2008Reading.pdf

Not enough of it was found to know the dimensions but an analysis of the lading combined with the extant hull fragments and info from other ships were enough to extrapolate some interesting data.
https://core.ac.uk/download/pdf/10126763.pdf

Carrying capacity: approx. 20 tons.
Length: 15.9 m
Beam (width): 5.3 m
Depth: 6.6 m
Draft 1 m
Freeboard: 5.6 m
Surface area: 103.5 m^2
cedar planking and keel: 4957.5 kg.
mast assembly: 930 kg
decks and beams: 582.4 kg
Total weight of ship: 6469.9 kg
Total weight including cargo and 6 crew: 28,366.68 kg

The final displacement of ship, cargo and crew is 28,726.68 kg at a draft of 1 m.

The "Rhinoceros" program they used looks like it might be pretty useful for creating ship models for gaming and other simulations.

[mlangsdorf]

Originally Posted by Rupert

No, it wouldn't. At about four times the same volume it might simply be four times as high from keel to top of its superstructure (having seen the way modern cruise ships are built this seems likely - I'd hate to try and pilot one in heavy winds). Even if you simply scale it up it'll be more like 1.54 times as big in each dimension, and while that would give it a shallower draft, that shallower draft in proportion to length and width arguably changes its lines (shallow ships, for the same displacement, tend to take more power to reach a given speed, assuming displacement hulls of similar design).

I'm using the GURPS terms for hydrodynamic lines here, not the naval engineering terms. For GURPS Vehicles, a Flower class corvette, a German model 1934 destroyer (Zerstorer 1934), an Illustrious class aircraft carrier, and an Iowa class battleship all have fine hydrodynamic lines, though the exact shapes of their hulls differ quite a bit and a Flower class corvette does not have the hull of an Iowa, scaled down by a factor of 3 or 4.

The example of the cube from the website is relevant here: a cube has no hydrodynamic lines and doesn't change its shape as you change its density, but a low density cube will have a shallower draft than a high density cube of the mass the same mass and less volume. In VE2, flotation rating is proportional to density, and so I really think that flotation rating should have something to do with a vehicle's draft.

[whswhs]

Originally Posted by DanHoward

Carrying capacity: approx. 20 tons.
Length: 15.9 m
Beam (width): 5.3 m
Depth: 6.6 m
Draft 1 m
Freeboard: 5.6 m
Surface area: 103.5 m^2
cedar planking and keel weigh 4957.5 kg.
mast assembly is 930 kg
decks and beams 582.4 kg
Total weight of ship: 6469.9 kg
Total weight including cargo and 6 crew: 28,366.68 kg

The final displacement of ship, cargo and crew is 28,726.68 kg at a draft of 1 m.

Hmmm. That works out to about 48 kg of planking and keel per square meter of surface. Cedar has a density about 60% that of water, so call it equivalent to 80 kg of water per square meter. Since a cubic meter of water is 1000 kg, we have a mean thickness of 8 cm, or just over 3 inches.

With a density of 37.5 lbs. per cubic foot, we have around 9.4 lbs. per square foot of hull. If that were all armor, and if cedar counts as expensive wood armor, we'd have about DR 10. I'm guessing, though, that the keel is much thicker and accounts for a bigger share of the weight, which would probably have to be classified as structure rather than armor.

[Polydamas]

Originally Posted by whswhs

I had been going for 7.5 cm, but that's still about double the thickness you cite. It also sounds as if we're looking at a ship with a super-light frame, not a frameless ship. The cited density of Aleppo pine is close enough to half that of water to make a 50% multiplier seem a reasonable figure, and the articles on the Kyrenia ship suggest that it's almost exactly the size I was aiming at, at least in length.

I don't know ships like this very well, but it seems like using a Super-Light or Extra-Light Frame to represent the cords and any keel or cross-timbers would be reasonable. I think that the hulls of many modern armoured vehicles are cast in a few pieces and welded together, and I think they are usually modelled as having a frame.

Yeah, the Kyrenia and Uluburun ships are about the size of your players' ships, even if the details are probably different.

Originally Posted by DanHoward

The Ulu Burun shipwreck is the best example of a TL1 trading ship.
http://cgs.la.psu.edu/documents/Pulak2008Reading.pdf

Thanks for taking the time to look up some stats, I could not spare it! It seems like their reconstruction is heavier than we are estimating the Kyrenia ship was.

[Agemegos]

Originally Posted by whswhs

Is this providing a formula for estimating the volume of a slice through an ellipsoid that extends from one of its poles to a plane of constant latitude? I can see that that would be useful in estimating the draft of a ship of known burden, by estimating the displacement volume needed to float that burden. It looks as if my copy of Hudson's Manual has a formula for that for a sphere, but the print's too small for the current state of my eyes; I'll have to take a look later under brighter light, and possibly with a magnifying glass.

There’s a formula in the article on “spherical cap” in Wikipedia. It’s the volume of the figure contained between a plane and a sphere, in terms of the radius of the sphere and the “height” of the cap (which amounts to the draught of a partly-submerged sphere).

[whswhs]

Originally Posted by Agemegos

There’s a formula in the article on “spherical cap” in Wikipedia. It’s the volume of the figure contained between a plane and a sphere, in terms of the radius of the sphere and the “height” of the cap (which amounts to the draught of a partly-submerged sphere).

Yes, I've looked at that, now, and it looks as if Hudson's Manual has the same formula, though it's explained somewhat less clearly. I'm guessing that what you have in mind is along the lines of "X% of the volume of this sphere is submerged, and that translates to Y% of the height, and that same percentage of the height of a partially submerged ellipsoid is a good estimate"?

I was also able to find a formula for estimating the surface area of an ellipsoid. When I plugged in the dimensions of a hypothetical ship, which I modeled as a fairly extreme ellipsoid, I came up with 355 square feet for the top deck, and 625 square feet for the hull, which looks tolerably close to the top deck being one-third of the area of the craft. That makes me feel a little better about using the Vehicles approximations, though I'm still leaning toward using the ellipsoid approximation instead of the cube approximation that much of Vehicles relies on.

[Agemegos]

If you have a sphere of radius r floating in liquid to depth h, then the volume of liquid that it displaces is V = πh²(3r-h)/3. Therefore it seems to me from linearity and similarity considerations that if you have an ellipsoidal hull with half-length a, half-length b, and half-height c floating to draught h in the intended orientation (with two principle axes parallel to the liquid surface and one perpendicular to it), that the hull must displace

V = ab/c² ∙ πh²(3c-h)/3

= π abh²(3c-h)/3c²

Translating to length A and beam B, and with c the height from keel to gunwale, I get

V = π ABh²(3c-h)/12c²

I don't feel like solving that cubic in closed form, but the solver tools in Excel ought to handle it.